Advanced Placement Chemistry: 1988 Free Response Answers





[delta] and [sigma] are used to indicate the capital Greek letters.
[square root] applies to the numbers enclosed in parenthesis immediately following
All simplifying assumptions are justified within 5%.
One point deduction for a significant figure or math error, applied only once per problem.
No credit earned for numerical answer without justification.
1) Average score = 4.95
a) four points
(1) 89.7 g SbCl5 / 299.0 g mole = 0.300 mole SbCl5
[SbCl5] initial = 0.300 mole / 15.0 liter = 0.200 M
(2) T = 182 °C + 273 = 455 K
P = nRT / V = [(0.300 mole) (0.0821 L atm) (455 K)] / [(15.0 L) (mole K)] = 0.747 atm
OR
(15.0 L / 0.300 mol) x (273 K / 455 K) = 30.0 L / mol (at std. temperature)
1 atm x [(22.4 L / mol) ÷ (30.0 L / mol)] = 0.747 atm
b) three points
Equilibrium concentrations:
[SbCl3] = [Cl2] = (0.0200 mol / L) x 0.292 = 5.84 x 10¯3 M
[SbCl5] = (0.0200 mol L) x 0.708 L = 1.42 x 10¯2 M
Kc = ([SbCl3][Cl2]) ÷ [SbCl5] = (5.84 x 10¯3)2 ÷ (1.42 x 10¯2) = 2.41 x 10¯3
OR
Equilibrium pressures:
PSbCl3 = PCl2
= 0.747 atm x 0.292 = 0.218 atm
PSbCl5 = 0.747 atm x 0.708 = 0.529 atm
Kp = (PSbCl3 x PCl2) ÷ PSbCl5 = (0.218)2 ÷ 0.529 = 8.98 x 10¯2
c) two points
K = ([SbCl3][Cl2]) ¯ [SbCl5] = 0.117
Equilibrium concentrations:
[SbCl5] = (1.00 - 0.70) mol / 2.00 L = 0.15 M
[SbCl3] = 0.700 mole / 2.00L = 0.350 M
[Cl2] = x
Kc = [(0.350) (x)] ÷ (0.15) = 0.117
x = [Cl2] = 0.50 M
Moles Cl2 at equilibrium = 0.050 mol L x 2.00 L = 0.10 mol
Moles Cl2 needed to make 0.300 mol SbCl3 into SbCl5 = 0.30 mol
Moles Cl2 that must be added = 0.40 mol
2) Average score = 3.25
a) two points
C + O2 --> CO2
2 H2 + O2 --> 2 H2O
C2H5OH --> 2CO2 + 3 H2O
(Half credit for recognition that combustion is reaction with O2)
b) three points
2 C + 2 O2 ---> 2 CO2
2 H2 + O2 ---> 2 H2O
2 CO2 + 3 O2 ---> C2H5OH + 3 O2
[delta]H° = 2 (- 393.5) = - 787.0 kJ
[delta]H° = 2(- 285.8) = - 571.6 kJ
[delta]H° = - (- 1366.7) = 1.366.7 kJ
Sum of three equations above
2 C + 2 H2 + H2O ---> C2H5OH
[delta]H° = - 8.1 kJ
(1 point for correct [delta]H° for each of the first 3 reations)
OR
[delta]H° combustion C(s) = [delta]Hf° CO2(g)
[delta]H° combustion H2(g) = [delta]Hf° H2O(l)
C2H5OH + 3 O2 ---> 2 CO2 + 3 H2O [delta]H° = -1,366.7kJ
[delta]H° reaction = [sigma] [delta]Hf° (products) - [sigma] [delta]Hf° (reactants)
= [2( -393.5) + 3(- 258.8) - [delta]Hf°C2H5OH - 0] kJ = - 277.7 kJ
2 C + 2 H2 + H2O ---> C2H5OH [delta]H° reaction = [sigma] [delta]H° (products) - [sigma] [delta]H°
(reactants)
= [- 277.7 - 0 - 0 - (- 285.8)] kJ = + 8.1 kJ
c) one point
[delta]S° = [sigma] S° (products) - [sigma] S° (reactants)
= (160.7 - 11.5 - 261.2 - 69.9) J / mol K = - 181.9 J / mol K
d) three points
(1) [delta]G° = [delta]H° - T [delta]S°
= 8,100 J - (298) (- 181.9) J = 8,100 J + 54,200 J = 62,300 J
(2) [delta]G° = - RTlnK
lnK = - [delta]G / RT
= - 62.300J ÷ [(8.31J/mol-K) (298K)] = - 25.2
K = 1.1 x 10¯11
OR
[delta]G° = - 2.303 RT logK
logK = - [delta]G / 2.303 RT
= - 62.000J ÷ [(2.303) (8.31) (298)] = - 10.9
K = 1.3 x 10¯11
If the terms are not rounded, K = 1.2 x 10¯11
correct substitutions in (1) and (2) earns one point
3) Average score = 5.02
a) two points
Sn ---> Sn2+ + 2e¯
Ag+ + e¯ ---> Ag
b) two points
2 Ag+ + Sn ---> 2 Ag + Sn2+
E° = [0.80 - (- 0.14)] V = 0.94 V
c) two points
E = (0.0591 ÷ n) log K (or - nFE = - RT ln K)
log K = (0.94 x 2) ÷ 0.0591 = 31.8
K = 6 x 1031
d) two points
E = E° - (0.0591 ÷ n) log [Sn2+] / [Ag+]2
OR
E = E° - (RT / nF) ln Q
Q = [Sn2+] / [Ag+]2
E = 0.94 = (0.0591 ÷ 2) log (1 ÷ (0.022))
E = 0.94 - 0.10 = 0.84 V
The final point is for completion of calculations in (c) and (d).
4) Average score = 5.66
3 points for each reaction; 1 pt. for reactants and 2 for products. Two points (of the three) for correct
response in inappropriate form, i.e. molecular when ionic is correct. One point penalty for spurious
product, e.g. redox product in an acid-base reaction. One point penalty for inclusion of spectator ions.
a) H+ (or H3O+) + I¯ + Cr2O72¯ ---> I2 + Cr3+ H2O
(Any reasonable I¯ oxidation product accepted.)
b) OH¯ + NH4+ ---> H2O + NH3 (or NH4OH)
c) Mg + Ag+ ---> Mg2+ + Ag
d) KClO3 ---> KCl + O2
(MnO2 is also shown over the reaction arrow in the original.)
e) H+ + CO32¯ ---> CO2 + H2O (or H2CO3 or HCO3¯)
f) SO3 + H2O ---> H+ + HSO4¯ (or H+ + SO42¯)
g) Ba2+ + SO42¯ ---> BaSO4
(or Ba2+ + HSO4¯ ---> BaSO4 + H+)
h) Cu2+ + NH3 ---> Cu(NH3)42+
(Partial credit for other logical Cu(II) ammonia complexes and for Cu(OH)2)
5) Average score = 2.99
a) two points
Xe and Ne are monatomic elements held together by London dispersion (van der Waals) forces. The
magnitude of such forces is determined by the number of electrons in the atom. A Xe atom has more
electrons than a neon atom has. (Size of the atom was accepted but mass was not.)
b) two points
The electrical conductivity of copper metal is based on mobile valence electrons (partially filled bands).
Copper chloride is a rigid ionic solid with the valence electrons of copper localized in individual
copper(II) ions.
c) two points
SiO2 is a covalent network solid. There are strong bonds many of which must be broken simultaneously to
volatize SiO2. CO2 is composed of discrete, nonpolar CO2 molecules so that the only forces holding the
molecules together are the weak London dispersion (van der Waals) forces.
d) two points
A lone pair of electrons on the central atom results in a pyramidal shape. The dipoles don't cancel, thus
the molecule is polar.
There is no lone pair on the central atom so the molecule has a trigonal planar shape in which the dipoles
cancel, thus the molecule is nonpolar.
6) Average score = 4.31
a) two points
The equilibrium pressure of NH3 gas would be unaffected Kp = (PNH3) (PH2S). Thus the amount of solid
NH4HS present does not affect the equilibrium.
b) two points
The equilibrium pressure of NH3 gas would decrease. In order for the pressure equilibrium constant, Kp,
to remain constant, the equilibrium pressure of NH3 must decrease when the pressure of H2S is increased.
Kp = (PNH3) (PH2S)
(A complete explanation based on Le Chatelier's principle is also acceptable.)
c) two points
The mass of NH4HS increases. A decrease in volume causes the pressure of each gas to increase. To
maintain the value of the pressure equilibrium constant, Kp, the pressure of each of the gases must
decrease. That decrease realized by the formation of more solid NH4HS.
Kp = (PNH3) (PH2S)
(A complete explanation based on Le Chatelier's principle is also acceptable.)
d) two points
The mass of NH4HS decreases because the endothermic reaction absorbs heat and goes nearer to
completion (to the right) as the temperature increases. (One point was assigned for each correct prediction
and one point for each correct explanation.)
7) Average score = 1.76
a) two points
The sharp vertical rise in the pH on the pH-volume curve appears at the equivalence point (about 23 mL).
Because the acid is monoprotic, the number of moles of acid equals the number of moles of NaOH. That
number is the product of the exact volume and the molarity of the NaOH. The molarity of the acid is the
number of moles of the acid divided by 0.030 L, the volume of the acid.
b) two points
At the half-equivalence point (where the volume of the base added is exactly half its volume at the
equivalence point), the concentration [HX] of the weak acid equals the concentration [X¯] of its anion.
Thus, in the equilibrium expression, [H+] [X¯] / [HX] = Ka. Therefore, pH at the half-equivalence point
equals pKa
c) one point
Cresol red is the best indicator because its pKa (about 8) appears midway in the steep equivalence region.
This insures that at the equivalence point the maximum color change for the minimal change in the
volume of NaOH added is observed.
d) three points
8) Average score = 3.82
a) four points
One point each for:
correct identification of axes
diagram of correct shape
correct labeling of regions
correct position of normal boiling point
b) one point
The argon sublimes.
c) one point
The argon vaporizes.
d) two points
The liquid phase is less dense than the solid phase. Since the freezing point of argon is higher than the
triple point temperature, the solid-liquid equilibrium line slopes to the right with increasing pressure.
Thus, if a sample of liquid argon is compressed (pressure increased) at constant temperature, the liquid
becomes a solid. Because increasing pressure favors the denser phase, solid argon must be the denser
phase.
9) Average score = 3.31
a) two points
Equipment needed includes a thermometer, and a container for the reaction, preferably a container that
serves as a calorimeter, and volumetric glassware (graduated cylinder, pipet....)
b) two points
Measurements include the difference in temperatures between just before the start of the reaction and the
completion of the reaction, and amounts (volume, moles) of the acid and the base.
c) two points
Determination of heat (evolved or absorbed): The sum of the volumes (or masses) of the two solutions,
the change in temperature and the specific heat of water are multiplied together to determine the heat of
solution for the sample used, (q = m x Cp x [delta]T). For full credit, two of the three factors must be
identified.
Division of the calculated heat of neutralization by moles of water produced, or moles of H+, or moles of
OH¯, or moles of limiting reagent.
d) two points
Experimental errors: heat loss to the calorimeter wall, to air, to the thermometer; incomplete transfer of
acid base from graduated cylinder; spattering of some of the acid or base so that incomplete mixing
occurred....Experimenter errors: dirty glassware, spilled solution, misread volume or temperature....(One
point for each of two experimental errors: one point for an experimenter error if only one experimental
error is given.)